Question

In Fig. altitudes AD and CE of A ABC
intersect each other at the point P. Show
that:
(i)triangle AEP ≈ triangle CDP
(ii) triangle ABD ≈ triangle CBE
(iii) triangle AEP ≈ triangle ADB
(iv) ΔPDC ≈ ΔBEC

​

Answer 1

Answer:

Hey friend!

Here is your answer:

Given,

ABC is a Δ in which

AD ⊥ BC

CE ⊥ AB

Ans 1)  To prove Δ AEP ≈ Δ CDP

In Δ AEP and Δ CDP,

we have:

∠ AEP = ∠ CDP = 90°

and ∠ APE = ∠ CPD

⇒ Δ AEP ≈ Δ CDP by A-A similarity

Ans 2) To prove Δ ABD ≈ Δ CBE

In Δ ABD and Δ CBE,

we have:

∠ CEB = ∠ ADB = 90°

and ∠ CBE = ∠ ABD

⇒ Δ ABD ≈ Δ CBE by A-A similarity      

Ans 3) To prove Δ AEP ≈ Δ ADB

In Δ AEP ≈ Δ ADB,

we have:

∠ EAP = ∠ BAD

and ∠ AEP = ∠ ABD = 90°

⇒ Δ AEP ≈ Δ ADB y A-A similarity

Ans 4) To prove ΔPDC ≈ ΔBEC

In  ΔPDC and ΔBEC,

we have:

∠PDC = ∠BEC = 90°

and ∠DCP = ∠ ECB

⇒  ΔPDC ≈ ΔBEC by A-A similarity

Hence proved

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Answer 2

Given,

ABC is a Δ in which

AD ⊥ BC

CE ⊥ AB

Ans 1)  To prove Δ AEP ≈ Δ CDP

In Δ AEP and Δ CDP,

we have:

∠ AEP = ∠ CDP = 90°

and ∠ APE = ∠ CPD

⇒ Δ AEP ≈ Δ CDP by A-A similarity

Ans 2) To prove Δ ABD ≈ Δ CBE

In Δ ABD and Δ CBE,

we have:

∠ CEB = ∠ ADB = 90°

and ∠ CBE = ∠ ABD

⇒ Δ ABD ≈ Δ CBE by A-A similarity      

Ans 3) To prove Δ AEP ≈ Δ ADB

In Δ AEP ≈ Δ ADB,

we have:

∠ EAP = ∠ BAD

and ∠ AEP = ∠ ABD = 90°

⇒ Δ AEP ≈ Δ ADB y A-A similarity

Ans 4) To prove ΔPDC ≈ ΔBEC

In  ΔPDC and ΔBEC,

we have:

∠PDC = ∠BEC = 90°

and ∠DCP = ∠ ECB

⇒  ΔPDC ≈ ΔBEC by A-A similarity

Hence proved

✌️

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