Question

Find the value of x in each case: Given: Iso. ΔABC, HM ∥DG Find: x, m∠CAB, m∠CBA

Answer 1

Since ΔABC is isosceles, m∠CAB = m∠CBA

Let m∠CAB = m∠CBA = y.

As HM || DG, JK || AB and JA, KB are transversals.

Therefore, m∠CAB = m∠CJK and m∠CBA = m∠CKJ (Corresponding angles)

So, m∠CJK = m∠CKJ= y.

m∠CLM = m∠CFG = 8x (Corresponding angles)

Now, m∠CJI = 7x - 25 and m∠CKL = 8x - 42 (Exterior angle theorem)

Note that m∠CJK + m∠CJI = 180 and

                m∠CKJ + m∠CKL = 180 (Linear pairs)

Therefore, y + (7x - 25) = 180   (1)

                  y + (8x - 42) = 180   (2)

Subtracting (1) from (2), we get,

x - 17 = 0

x = 17 degrees

y + (7x - 25) = 180

y + [7(17) - 25] = 180

y + 119 - 25 = 180

y + 94 = 180

y = 180 - 94 = 86

Therefore, m∠CAB = m∠CBA = 86 degrees

Answer 2

1. Start with ΔCIJ.

• ∠HIC and ∠CIJ are supplementary, then m∠CIJ=180°-7x;
• the sum of the measures of all interior angles in ΔCIJ is 180°, then m∠CJI=180°-m∠JCI-m∠CIJ=180°-25°-(180°-7x)=7x-25°;
• ∠CJI and ∠KJA are congruent as vertical angles, then m∠KJA =m∠CJI=7x-25°.

2. Lines HM and DG are parallel, then ∠KJA and ∠JAB are consecutive interior angles, then m∠KJA+m∠JAB=180°. So

m∠JAB=180°-m∠KJA=180°-(7x-25°)=205°-7x.

3. Consider ΔCKL.

• ∠LFG and ∠CLM are corresponding angles, then m∠LFG=m∠CLM=8x;
• ∠CLM and ∠CLK are supplementary, then m∠CLM+m∠CLK=180°, m∠CLK=180°-8x;
• the sum of the measures of all interior angles in ΔCLK is 180°, then m∠CKL=180°-m∠CLK-m∠LCK=180°-(180°-8x)-42°=8x-42°;
• ∠CKL and ∠JKB are congruent as vertical angles, then m∠JKB =m∠CKL=8x-42°.

4. Lines HM and DG are parallel, then ∠JKB and ∠KBA are consecutive interior angles, then m∠JKB+m∠KBA=180°. So

m∠KBA=180°-m∠JKB=180°-(8x-42°)=222°-8x.

5. ΔABC is isosceles, then angles adjacent to the base are congruent:

m∠KBA=m∠JAB → 222°-8x=205°-7x,

7x-8x=205°-222°,

-x=-17°,

x=17°.

Then m∠CAB=m∠CBA=205°-7x=86°.

Answer: 86°.