# Similar Triangles Questions

## Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string ( from the tip of her rod to the fly) is taut, how much string does she have out (see fig.) ? If she pulls in the string at the rate of 5 cm per second, what will the horizontal distance of the fly from her after 12 seconds ?

Solution : Let O be the position of nazima at the time of fishing. Let OA be the rod and AB be the string. Height of the tip of rod above sea level. i.e.    AC = 1.8 m The distance of the fly from the girl = 3.6 m i.e.   OB = 3.6 m …

## In the figure, D is a point on side BC of triangle ABC such that $$BD\over CD$$ = $$AB\over AC$$. Prove that AD is the bisector of $$\angle$$ BAC.

Solution : Given : ABC is a triangle and D is point on BC such that $$BD\over CD$$ = $$AB\over AC$$ To Prove : AD is the bisector of $$\angle$$ BAC. Construction : Produce line BA to E such that line AE = AC. Join CE. Proof : In $$\triangle$$ AEC, since AE = AC, …

## In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (i) $$\triangle$$ PAC ~ $$\triangle$$ PDB (ii) PA.PB = PC.PD

Solution : (i)  In $$\triangle$$s PAC and PDB, we have : $$\angle$$ APC = $$\angle$$ DPB         (common) $$\angle$$ PAC = $$\angle$$ PDB [$$\therefore$$  $$\angle$$ BAC = 180 – $$\angle$$ PAC  and $$\angle$$ PDB = $$\angle$$ CDB = 180 – (180 – $$\angle$$ PAC) = $$\angle$$ PAC] $$\therefore$$  By AA similarity, we have …

## In the figure, two chords AB and CD intersect each other at the point P. Prove that (i) $$\triangle$$ APC ~ $$\triangle$$ DPB (ii) AP.PB = CP.DP

Solution : (i)  In $$\triangle$$s PAC and PDB, we have : $$\angle$$ APC = $$\angle$$ DPB         (vertically opp. angles) $$\angle$$ CAP = $$\angle$$ BDP        (angles in same segment of circle are equal) $$\therefore$$  By AA similarity, we have : $$\triangle$$ APC ~ $$\triangle$$ DPB  (ii)  Since $$\triangle$$s APC ~ …

## Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

Solution : We know that AD is a median of triangle ABC, then $${AB}^2$$ + $${AC}^2$$ = 2$${AD}^2$$ + $${1\over 2}{BC}^2$$ Since the diagonals of a parallelogram bisect each other, therefore BO and DO are medians of triangles ABC and ADC respectively. $$\therefore$$  $${AB}^2$$ + $${BC}^2$$ = 2$${BO}^2$$ + $${1\over 2}{AC}^2$$          …

## Prove that in any triangle, the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.

Solution : Given : ABC is a triangle and AD is the median. To Prove : (i)  $${AB}^2$$ + $${AC}^2$$ = 2[$${AD}^2$$ + $${BD}^2$$] (ii)  $${AB}^2$$ + $${AC}^2$$ = 2$${AD}^2$$ + 2$$[{1\over 2}BC]^2$$ Construction : Draw AE $$\perp$$ BC Proof : In right angled triangle ABE, by Pythagoras theorem, $${AB}^2$$ = $${BE}^2$$ + $${AE}^2$$    …

Solution : Given : In triangle ABC angle B is an acute angle and AD $$\perp$$ BC To Prove : $${AC}^2$$ = $${AB}^2$$ + $${BC}^2$$ – 2BC.BD Proof : Since $$\triangle$$ ADC is a right triangle, angled at D. Therefore, by Pythagoras theorem, $${AC}^2$$ = $${AD}^2$$ + $${CD}^2$$ $$\implies$$  $${AC}^2$$ = $${AD}^2$$ + $${BC – … ## In the figure, ABC is a triangle in which \(\angle$$ ABC > 90 and AD $$\perp$$ CB produced. Prove that $${AC}^2$$ = $${AB}^2$$ + $${BC}^2$$ + 2BC.BD.

Solution : Given : ABC is triangle in which $$\angle$$ ABC > 90 and AD $$\perp$$ CB produced. To Prove : $${AC}^2$$ = $${AB}^2$$ + $${BC}^2$$ + 2BC.BD Proof : Since $$\triangle$$ ADB is a right triangle, angled at D. Therefore, by Pythagoras theorem, $${AB}^2$$ = $${AD}^2$$ + $${DB}^2$$           ……..(1) Again, …

## In the figure, D is a point on hypotenuse AC of $$\triangle$$ ABC, BD $$\perp$$ AC, DM $$\perp$$ BC and DN $$\perp$$ AB. Prove that (i) $${DM}^2$$ = DN.MC (ii) $${DN}^2$$ = DM.AN

Solution : We have : AB $$\perp$$ BC and DM $$\perp$$ BC. So,        AB || DM Similarly, we have : BC $$\perp$$ AB and DN $$\perp$$ AB. So,          CB || DN Hence, quadrilateral BMDN is a rectangle. $$\therefore$$  BM = DN (i)  In triangle BMD, we have : …

## In the figure, PS is the bisector of $$\triangle$$ PQR. Prove that $$QS\over SR$$ = $$PQ\over PR$$.

Solution : Given : PQR is a triangle and PS is the bisector of $$\angle$$ QPR meeting QR at S. $$\therefore$$  $$\angle$$ QPS = $$\angle$$ SPR To Prove : $$QS\over SR$$ = $$PQ\over PR$$ Construction : Draw RT parallel to SP to cut QP produced at T. Proof : Since PS || TR and PR …