Q. 74.6( 231 Votes )

In Fig. 6.38, altitudes AD and CE of Δ ABCintersect each other at the point P. Showthat:

(i) Δ AEP ~ Δ CDP

(ii) Δ ABD ~ Δ CBE

(iii) Δ AEP ~ Δ ADB

(iv) Δ PDC ~ Δ BEC


Answer :

(i) In ΔAEP and ΔCDP,

AEP = CDP (Each 90°)


APE = CPD (Vertically opposite angles)


Hence, by using AA similarity,


ΔAEP ΔCDP


(ii) In ΔABD and ΔCBE,


ADB = CEB (Each 90°)


ABD = CBE (Common)


Hence, by using AA similarity,


ΔABD ΔCBE


(iii) In ΔAEP and ΔADB,


AEP = ADB (Each 90°)


PAE = DAB (Common)


Hence, by using AA similarity,


ΔAEP ΔADB


(iv) In ΔPDC and ΔBEC,


PDC = BEC (Each 90°)


PCD = BCE (Common angle)


Hence, by using AA similarity,


ΔPDC ΔBEC


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