Question

In Fig. 6.38, altitudes AD and CE of ∆ ABC intersect each other at the

point P. Show that:

(i) ∆AEP ~ ∆ CDP

(ii) ∆ABD ~ ∆ CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆ PDC ~ ∆ BEC​

Answer 1

Answer:

In two altitudes are and which intersect each other at point .

(i)

In ,

Hence, , by AA similarity criterion.

(ii)

In ,

Hence, , by AA similarity criterion.

(iii)

In ,

Hence, , by AA similarity criterion.

(iv)

In ,

Hence, , by AA similarity criterion.

Answer 2

Answer:

in triangle AEP and triangle CDP,

angle Aep =angle cdp=90°

angle ape =angle cpd(vertically opposite angle)

angle pae=angle pcd(remaining angle)

therefore by aaa similarity criteria,

triangle aep similar to triangle cdp.

similarly in the same way you have to prove other three also.