MHB Area of Isosceles Triangle proof

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The discussion focuses on understanding the derivation of the third line in a proof related to the area of an isosceles triangle. The process begins by multiplying the expression $h(p + q)$ by $ab/ab$ to maintain equality. This manipulation leads to a reformulation of the area expression, breaking it down into components involving the heights and bases of the triangle. The final expression simplifies to $\frac{1}{2}h(p + q)$, demonstrating how the area is calculated. Clarity on these steps is essential for grasping the proof's logic.
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I want to know how did it arrive at the third line of this proof. I didn't get it. Thanks!
 

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paulmdrdo said:
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

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

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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