In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that

(i) ΔAEP ~ ΔCDP

(ii) ΔABD ~ ΔCBE

(iv) ΔPDC ~ ΔBEC

Asked by Abhisek | 1 year ago |  69

##### Solution :-

Given, altitudes AD and CE of ΔABC intersect each other at the point P.

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

Hence, by AA similarity criterion,

ΔABD ~ ΔCBE

∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

Hence, by AA similarity criterion,

ΔPDC ~ ΔBEC

Answered by Pragya Singh | 1 year ago

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