Question 1. If two sides of one triangle are equal to the two sides of another triangle and the contained angles are supplementary; show that the two triangles are equal in area.
Let ∆ABC and ∆DEF are such that AC = DE; CB = EF and ∠ABC + ∠DEF = 180°.
Now, allow ∆DEF fall along ∆ABC such that DE falls on AC and B and F are on opposite directions.
∵∠ACB + ∆DEF = 180°
⇒ BCF is a straight line
∴ We get a ∆AFB, in which C is mid point of FB.
∴ AC is median of ∆AFB, Median divides the ∆ into two triangles equal in area.
⇒ ar (∆AFC) = ar (∆ABC)
⇒ ar (∆DEF) = ar (∆ABC).
Question 2. In a trapezium ABCD, where AB  DC, E is the mid point of BC. Prove that ∆AED = ½ trap. ABCD.
Hint: Through E, draw FEG  AD meeting AB and CD in F and G respectively.
∴∆ BEF ≅ ∆ CEG
⇒ ar trap. ABCD = ar (^{gm} ADGF)
⇒ ∆ AED = ½ ^{gm} ADGF
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