Question

Altitudes AD and CE of ∆ABC intersect each other at the point P. Show that :- (1) ∆AEP congruent to ∆CDP. (2)∆ABD congruent to ∆CBE (3)∆AEP congruent to to ∆ADB (4)∆PDC congruent to∆BEC​

Answer 1

Step-by-step explanation:

1). In ∆AEP&∆CDP

ANGLE APE=angle CPD(Vertically opposite angles)

AE=CD. (EQUAL angles have opposite sides equal)

angle AEP=angle PDC (EACH 90)

HENCE PROVED

2) IN ∆ABD&∆CBE

ANGLE. ADB= ANGLE CEB(EACH 90)

AB=CB(EQUAL angles have opposite sides equal)

ANGLE ABD= ANGLE EBC(COMMON)

HENCE PROVED

Answer 2

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Given that AD and CE are the altitudes of triangle ABC and these altitudes intersect each other at 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

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

Hence, by AA similarity criterion,

ΔPDC ~ ΔBEC

Hope it's Helpful....:)