Question

$\huge \colorbox{cyan}{QUESTION}$
In the given figure, altitudes AD and CE of ∆ ABC intersect each other at the point P. Show that:

(i) ∆AEP ~ ∆ CDP

(ii) ∆ABD ~ ∆ CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆ PDC ~ ∆ BEC


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Answer 1

Step-by-step explanation:

data: altitudes AD and CE of ABC intersect each other at the point P.

To prove:

(i) ∆AEP ~ ∆CDP

(ii) ∆ABD ~ ∆CBE

(iii) ∆AEP ~ ∆ADB

(iv) ∆PDC ~ ∆BEC

☈Solution:-

⬤(i) In ∆AEP and ∆CDP.

∠AEP = ∠CDP = 90° (data) .

∠APE = ∠CPD (vertically opposite angle)

∴ ∠PAE = ∠PCD these are equiangular triangles.

similarity criterion for ∆ is A.A.A

∴ ∆AEP ~ ∆CDP ✔

⬤ (ii) in ∆ABD and ∆CBE.

∠ADB = ∠CEB = 90° (data)

∠ABD = ∠CBE (common)

∴ ∠DAB = ∠BCE These are equiangular triangles.

∴ Similarity criterion for is A.A.A.

∴ ∆ABD ~ ∆CBE ✔

⬤(iii) In ∆AEP and ∆ADB.

∠AEB = ∠ADB = 90° (data)

∠PAE = ∠DAB (common)

∴ ∠APE = ∠ABD.

∴ these are equiangular triangles.

∴ similarity criterion for is A.A.A.

∴ ∆AEP ~ ∆ADB ✔

⬤(iv) In ∆PDC and ∆BEC.

∠PDC = ∠BEC = 90° (data)

∠PCD = ∠BCE (common)

∴ ∠CPD = ∠CBE

∴ These are equiangular angular triangles.

Similarity criterion for ∆ is A.A.A.

∴ ∆PDC ~ ∆BEC✔

I hope it's help you...☺

Answer 2

Answer:

$\huge \fcolorbox{purple}{lavenderblush}{hello}$

yes sure ❣️