MHB Area of Isosceles Triangle proof

[paulmdrdo1]

I want to know how did it arrive at the third line of this proof. I didn't get it. Thanks!

 

[Euge]

I want to know how did it arrive at the third line of this proof. I didn't get it. Thanks!

Hi paulmdrdo,

First multiply $h(p + q)$ by $ab/ab$ (which equals $1$) to get the expression

$$ab \cdot \frac{h(p + q)}{ab}.$$

Since

$$ \frac{h(p + q)}{ab} = \frac{hp+hq}{ab} = \frac{p}{a}\frac{h}{b} + \frac{h}{a}\frac{q}{b},$$

we have

$$ ab \cdot \frac{h(p + q)}{ab} = ab\left(\frac{p}{a}\frac{h}{b} + \frac{h}{a}\frac{q}{b}\right)$$

and therefore

$$\frac{1}{2}h(p + q) = \frac{1}{2}ab\left(\frac{p}{a}\frac{h}{b} + \frac{h}{a}\frac{q}{b}\right).$$