In the figure, altitudes AD and CE of \(\triangle\) ABC intersect each other at the point P. Show that : (i) \(\triangle\) AEP ~ \(\triangle\) CDP (ii) \(\triangle\) ABD ~ \(\triangle\) CBE (iii) \(\triangle\) AEP ~ \(\triangle\) ADB (iv) \(\triangle\) PDC ~ \(\triangle\) BEC

Solution :

(i)  In triangle\(\triangle\) AEP and CDP, we have

\(\angle\) AEP = \(\angle\) CDP = 90

\(\angle\) APE = \(\angle\) CPD           (vertically opposite angles)

\(\therefore\)  By AA similarity, we have :

\(\triangle\) AEP ~ \(\triangle\) CDP

(ii)  In \(\triangle\) ABD and CBE, we have

\(\angle\) ABD = \(\angle\) CBE         (common angle)

\(\angle\) ADB = \(\angle\) CEB = 90

\(\therefore\)  By AA similarity, we have :

\(\triangle\) ABD ~ \(\triangle\) CBE

(iii)  In \(\triangle\) AEP and ADB, we have

\(\angle\) AEP = \(\angle\) ADB = 90

\(\angle\) PAE = \(\angle\)  DAB          (common angles)

\(\therefore\)  By AA similarity, we have :

\(\triangle\) AEP ~ \(\triangle\) ADB

(iv)  In \(\triangle\) PDC and BEC, we have

\(\angle\) PDC = \(\angle\) BEC = 90

\(\angle\) PCD = \(\angle\) ECB           (common angles)

\(\therefore\)  By AA similarity, we have :

\(\triangle\) PDC ~ \(\triangle\) BEC

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