Question : In $\triangle$ABC, the bisector of $\angle$BAC intersects BC at D and the circumcircle of $\triangle$ABC at E. If AB : AD = 3 : 5, then AE : AC is:
Option 1: 5 : 3
Option 2: 3 : 2
Option 3: 2 : 3
Option 4: 3 : 5
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: 3 : 5
Solution : Given: AB : AD = 3 : 5 AE is bisector of $\angle$BAC. So, $\angle$BAD = $\angle$CAE $\angle$AEC = $\angle$ABC [As AC is a common segment] $\therefore \triangle$ABD $\sim \triangle$AEC So, $\frac{AB}{AE}=\frac{AD}{AC}$ ⇒ $\frac{AB}{AD}=\frac{AE}{AC}$ $\therefore \frac{AE}{AC}=\frac{3}{5}$ Hence, the correct answer is 3 : 5.
Candidates can download this ebook to know all about SSC CGL.
Admit Card | Eligibility | Application | Selection Process | Preparation Tips | Result | Answer Key
Question : In triangle ABC, the bisector of angle BAC cuts the side BC at D. If AB = 10 cm, and AC = 14 cm then what is BD : BC ?
Question : In triangle ABC, the bisector of angle BAC cuts the side BC at D. If AB = 10 cm, and AC = 14 cm, then what is BD : DC?
Question : In a triangle ABC a straight line parallel to BC intersects AB and AC at D and E, respectively. If AB = 2AD, then DE : BC is:
Question : ABC is a right angle triangle and $\angle ABC = 90^{\circ}$. BD is perpendicular on the side AC. What is the value of $(BD)^2$?
Question : In $\triangle$ABC, $\angle$B = 35°, $\angle$C = 65° and the bisector of $\angle$BAC meets BC in D. Then $\angle$ADB is:
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile