# Two Triangles

Prove that the angle marked $a$ is half the size of the angle marked $b$.

In the diagram, triangle ABC is isosceles with AB = BC.

The line AP is drawn so that it meets the side BC at a right angle.

Prove that the angle marked $a$ is half the size of the angle marked $b$.

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**Using the isosceles triangle and the right-angled triangle**

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”‹From the right-angled triangle, $a+c+90=180$, so $a=90-c$.

From the isosceles triangle, $2c+b=180$, so $b=180-2c$.

But $180-2c=2(90-c)$, so $b=2a$.

Using this pair of triangles can also lead to simple simultaneous equations.

**Using the two right-angled triangles**

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From the blue right-angled triangle, $a+c+90=180$, so $c=90-a$.

Similarly, the missing angle in the green right-angled triangle is $90-b$.

But the missing angle in the green triangle added to $a$ makes $c$.

So $a+90-b=90-a$, which rearranges to give $2a=b$.