Question

Question 7 In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ∼ ΔCDP (ii) ΔABD ∼ ΔCBE (iii) ΔAEP ∼ ΔADB (v) ΔPDC ∼ ΔBEC

Class 10 - Math - Triangles Page 140

Answer 1

(i) In ΔAEP and ΔCDP,
∠AEP = ∠CDP (Each 90°)
∠APE = ∠CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,
∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ~ ΔCBE


(iii) In ΔAEP and ΔADB,
∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)
Hence, by using AA similarity criterion,
ΔAEP ~ ΔADB



(iv) In ΔPDC and ΔBEC,
∠PDC = ∠BEC (Each 90°)
∠PCD = ∠BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ~ ΔBEC

Answer 2

Answer:

It's in the attachment

Follow me and mark me as a brainliest