Chapter 5.1, Problem 39WE

To determine

To prove: that lengths of the segments drawn from any point are the legs equal to length of altitude drawn to one leg.

Answer to Problem 39WE

The lengths of the segments drawn from any point are the legs equal to length of altitude drawn to one leg is proved as:

  $\begin{array}{cc}BX+CX& =BC\\ kX+kXY& =kCD\\ XZ+XY& =CD\end{array}$

Explanation of Solution

Given information:

The sum of lengths of the segments drawn from any point in the base of an isosceles triangle perpendicular to the legs is equal to length of altitude drawn to one leg.

proof:

Consider that above information:

Given: Triangle $ABC$ with $AB=AC$

The point $D$ is on $AB$ with $CD$ perpendicular to $AB.X$ is between $B$ and $C.Y$ is on $AC$ with $XY$ perpendicular to $AC;Z$ is on $AB$ with $XZ$ perpendicular to $AB$

Consider the figure given below:

  

Prove: $XZ+XY=CD$

Proof: By the Isosceles Triangle Theorem, angles $ABC$ and $ACB$ are congruent. The following three triangles are similar (all by Angle Angle): $BZX,BDC$ , and $CYX$

It can be prove as,

  $BX=CX$

  $BZ=YC$

  $ZX=YX$

Hence, the triangles are congruent $\text{Δ}BZX\cong \text{Δ}CYX$ .

For similar triangle, Ratio is same

  $\frac{BX}{XZ}=\frac{CX}{XY}=\frac{BC}{CD}=k$

Then use the fact that $BX+CX=BC$

  $\begin{array}{cc}BX+CX& =BC\\ kX+kXY& =kCD\\ XZ+XY& =CD\end{array}$

Hence the above is proved.