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Two Triangles Having the Same Base & Equal Areas

The median AA...

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The median AA' of the \bigtriangleup ABC meets the side B'C' of the medial triangle A'B'C' in P and CP meets AB in Q .Show that in the triangle ABC,AB=3A

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Q. In triangle ABC ,D is the midpoint of AB. P IS ANY POINT OF BC. CQ || PD meets AB in Q . Show that ar ( ∆BPQ) = ½(∆ABC)

In triangle ABC; $∠$ A = $60^{0}$ , $∠$ C = $40^{0}$ and bisector of angle ABC meets side AC at point P. Show that BP= CP.

△ A B C, D

is the mid-point of

A B

P

is any point on

B C

C Q | | P D

A B

Q

, then prove that area

(△ B P Q) = \frac{1}{2}

(△ A B C)

Q. In triangle ABC, D is mid-point of AB and P is any point on BC. If CA parallel to PD meets AB at Q, prove that:
2× area(∆ABC)= area (∆ABC)

△ A B C

A B C

is equal to twice the angle

A C B

, and bisector of angle

A B C

meets the opposite side at point

P

, show that:
(i)

C B : B A = C P : P A

A B \times B C = B P \times C A

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Two Triangles Having the Same Base & Equal Areas

Standard IX Mathematics

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