Triangle ABC and DBC are on the same base BC with A and D on opposite sides of line BC such that area $(Δ A B C)$ = area $(Δ D B C)$ . Show that BC bisects AD.

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Solution

Given: Figure $A B C$ and $D B C$ are on the same base $B C$ with vertices $A$ and $D$ or opposite sides of $B C$ such that $a r (△ A B C) = a r (△ D B C)$
To Prove: $B C$ bisects $A D$
Proof: $a r (△ A B C) = a r (△ D B C)$
$\Rightarrow \frac{1}{2} \times B C \times A M$
$\Rightarrow \frac{1}{2} \times B C \times D N$
Area of triangle $= \frac{1}{2} \times$ Base $\times$ Corresponding altitude
$\Rightarrow A M = D N - - - - - (1)$
In $△ A M D$ and $△ D N O$
$A M = D N$ (prove (1))
$∠ A M N = ∠ D N O$ (each $90^{o}$ )
$∠ A O M = ∠ D O N$
vertically opposite angles
$∴ △ A M O ⊥ △ D N O$
are to $A A S$ congurence rule
$∴ A D = D O (C P C T)$
$\Rightarrow B C$ bisects $A D$

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Triangle ABC and DBC are on the same base BC with A and D on opposite sides of line BC such that area (ΔABC)= area (ΔDBC). Show that BC bisects AD.

Triangle ABC and DBC are on the same base BC with A and D on opposite sides of line BC such that area $(Δ A B C)$ = area $(Δ D B C)$ . Show that BC bisects AD.