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Isosceles Triangle Property

Prove that ar...

Question

Prove that $a r (Δ A P B) = a r (Δ A C Q)$

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Solution

In a $Δ A X P$ and $Δ C X B$ ,
$∠ P A X = X C B$ (alternative angles AP||BC)
AX=CX(given)
$∠ A X P = ∠ C X B$ (vertically opposite angles)
$Δ ≃ Δ C X B$ (ASA rule)
$\Rightarrow A P = B C$ (By cpct) .....(1)
Similarly QA=BC .....(2)
From (1) and (2) we get
$A P = Q A$
Now, AP||BC and AP=QA
$a r (Δ A P B) = a r (Δ A C Q)$ [Triangles having equal bases and between the same parallels QP and BC].

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Q. In the give figure, X and Y are the midpoints of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar(∆ABP) = ar(∆ACQ).

Q. Question 9
In figure X and Y are the mid - points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that

a r (Δ A B P) = a r (Δ A C Q)

Q. Question 9
In figure X and Y are the mid - points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that

a r (Δ A B P) = a r (Δ A C Q)

In parallelogram ABCD, AP and CQ are perpendicular to diagonal BD.
It can be proved that $Δ$ APB $≅$ $Δ$ CQD.

Q. In the adjacent figure

A B C D

is a square and

Δ A P B

is an equilateral triangle. Prove that

Δ A P D ≅ Δ B P C

Δ A P D

Δ B P C ¯ ¯¯¯¯¯¯¯ ¯ A D = ¯ ¯¯¯¯¯¯ ¯ B C, ¯ ¯¯¯¯¯¯ ¯ A P = ¯ ¯¯¯¯¯¯ ¯ B P

∠ P A D = ∠ P B C = 90^{\circ} - 60^{\circ} = 30^{\circ})

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