In the following figure, ABCD is a parallelogram. Prove that:angleDAP + angleCBP= angleAPB | 9 |…
In the following figure, ABCD is a parallelogram. Prove that:angleDAP + angleCBP= angleAPB | 9 |…

In the given figure, $ABCD$ is a parallelogram then prove that $\triangle ABE=\triangle BCE$

Attempt $$\triangle ABD=\triangle CBD$$ $$\triangle ABD=\frac {1}{2}||gm ABCD$$

Now what should I do next?

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In the given figure, $ABCD$ is a parallelogram then prove that $\triangle ABE=\triangle BCE$

Attempt $$\triangle ABD=\triangle CBD$$ $$\triangle ABD=\frac {1}{2}||gm ABCD$$

Now what should I do next?

Draw a line AP and a line CQ that are perpendicular to DE.

$\triangle ADE=\frac{AP\times DE}{2}$

$\triangle CDE=\frac{CQ\times DE}{2}$

$\triangle BCE=\frac{AP\times DE}{2}-\triangle CDB$

$\triangle ABE=\frac{CQ\times DE}{2}-\triangle ADB$

Since ABCD is parallelogram, $\triangle CDB=\triangle ADB$

Since $\triangle ADP$ is congruent with $\triangle CBQ$, $AP=CQ$.

Therefore, $\triangle BCE=\triangle ABE$.

Slightly easier. Connect $AC$. Let the intersection of the two diagonals $AC$ and $BD$ be $M$.$\triangle MAD \equiv \triangle MCB$ is easily shown. Hence the altitudes of these triangles to the bases $MD$ and $MB$ are equal. $\triangle ABE$ shares the same altitude as $\triangle MAD$ and $\triangle BCE$ shares the same altitude as $\triangle MCB$ with respect to the common base $BE$. Since $\triangle ABE$ and $\triangle BCE$ have equal altitudes and a common base, they have equal areas.

Consider the triangle $\triangle ADE$. The segment $AB$ divides $\triangle ADE$ into two triangles, $\triangle ABD$ and $\triangle ABE$, whose areas are in the ratio $$\lvert\triangle ABD \rvert : \lvert\triangle ABE \rvert = BD : BE.$$ This is a fact whenever you subdivide a triangle in this way, and it is a consequence of the fact that all three triangles $\triangle ADE$, $\triangle ABD$, and $\triangle ABE$ have bases along the line $DE$ and have the same altitude from vertex $A$ to those bases.

A similar argument regarding the triangle $\triangle CDE$ tells us that $$\lvert\triangle CBD \rvert : \lvert\triangle CBE \rvert = BD : BE.$$

Two ratios that are equal to $BD:BE$ are equal to each other, so $$\lvert\triangle ABD \rvert : \lvert\triangle ABE \rvert = \lvert\triangle CBD \rvert : \lvert\triangle CBE \rvert.$$

This implies $$\lvert\triangle ABE \rvert : \lvert\triangle CBE \rvert = \lvert\triangle ABD \rvert : \lvert\triangle CBD \rvert.$$

But since $\lvert\triangle ABD \rvert = \lvert\triangle CBD \rvert$, this implies that $\lvert\triangle ABE \rvert = \lvert\triangle CBE \rvert$.

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