# 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

(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,

ABD = CBE (Common)

Hence, by using AA similarity,

ΔABD ΔCBE

PAE = DAB (Common)

Hence, by using AA similarity,

(iv) In ΔPDC and ΔBEC,

PDC = BEC (Each 90°)

PCD = BCE (Common angle)

Hence, by using AA similarity,

ΔPDC ΔBEC

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