In \(\triangle ABC\), points D and E are on AB and AC respectively such that \(AD=2\,DB\) and \(AE=2\,EC\). If \(\angle DAE=95^\circ\) and \(\angle ADE=42^\circ\), then find \(\angle ACB\).
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(43^\circ\)
By the converse of the basic proportionality theorem, a line that divides two sides of a triangle in the same ratio is parallel to the third side. If DE is parallel to BC, the angles at E and C are corresponding angles and are equal.
Showing that DE is parallel to BC:
\(AD=2\,DB\), so \(\displaystyle \frac{AD}{DB}=2\)
\(AE=2\,EC\), so \(\displaystyle \frac{AE}{EC}=2\)
The ratios are equal, so DE \(\parallel\) BC
Finding the angle:
In \(\triangle ADE\): \(\angle AED=180^\circ-95^\circ-42^\circ=43^\circ\)
\(\angle ACB=\angle AED\) (corresponding angles, DE \(\parallel\) BC), so \(\angle ACB=43^\circ\)