Proofs in Plane Geometry practice questions

• $\text{Join } AC \text{ and let } G \text{ be the midpoint of } AC\text{.}$
• $\text{In } \triangle ACD\text{, } E \text{ and } G \text{ are the midpoints of } AD \text{ and } AC\text{.}$
• $EG \text{ is parallel to } DC \text{ and } EG = \frac{1}{2} DC \quad \scriptsize\textit{midpoint theorem}$
• $\text{In } \triangle ABC\text{, } G \text{ and } F \text{ are the midpoints of } AC \text{ and } BC\text{.}$
• $GF \text{ is parallel to } AB \text{ and } GF = \frac{1}{2} AB \quad \scriptsize\textit{midpoint theorem}$
• $\text{As } AB \text{ is parallel to } DC\text{, } EG \text{ is parallel to } GF\text{.}$
• $\text{They meet at } G\text{, so } EGF \text{ is a straight line.}$
• $EF = EG + GF$

Answer: $12$

• $\text{Since } DE \text{ is parallel to } BC, \quad \triangle ADE \text{ is similar to } \triangle ABC \quad \scriptsize\textit{(AA similarity)}$
• $\frac{DE}{BC} = \frac{AD}{AB}$
• $= \frac{3}{17}$
• $\frac{12}{BC} = \frac{3}{17}$
• $BC = \frac{12 \times 17}{3}$
• $= 68 \text{ cm}$

Answer: $68$

• $\text{By the Tangent-Chord Theorem, } \angle ACB = \angle TAB$
• $= 44^\circ$
• $\text{In } \triangle ABC: \quad \angle BAC + \angle ABC + \angle ACB = 180^\circ$
• $\angle BAC = 180^\circ - 34^\circ - 44^\circ$
• $= 102^\circ$

Answer: $102$