   Chapter 5.6, Problem 39E Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Solutions

Chapter
Section Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

Given: Δ A B C (not shown) is isosceles with m ∠ A B C = m ∠ C = 72 ∘ ; B D → bisects ∠ A B C and A B = 1 Find: B C

To determine

To find:

BC.

Explanation

Given:

In an isosceles ΔABC with mABC=mC=72, BD bisects ABC and AB=1.

Theorem used:

Angle bisector theorem:

If a ray bisects one angle of a triangle, then it divides the opposite side into segments whose lengths are proportional to the lengths of the two sides that form the bisected angle.

Calculation:

Consider an isosceles ΔABC.

We have mABC=mC=72.

Since BD bisects ABC, mABD=mDBC=12ABC.

mABD=12ABC=12(72)=36

We know that the sum of angles in a triangle is 180.

mA+mB+mC=180mA+72+72=180mA+144=180mA=180144mA=36

Thus mABD=mA=36

ΔABD is an isosceles triangle, with AD=DB.

Consider the ΔBDC.

mBDC+mDBC+mDCB=180mBDC+36+72=180mBDC+108=180mBDC=180108mBDC=72

That is, mBDC=mDCB

Hence ΔBDC is an isosceles triangle with BD=BC

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