Question : In $\Delta ABC,\angle BAC=90^{\circ}$ and D is the mid-point of BC. Then which of the following relations is true?
Option 1: $AD=BD=CD$
Option 2: $AD=BD=2CD$
Option 3: $AD=2BD=CD$
Option 4: $2AD=BD=CD$
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Correct Answer: $AD=BD=CD$
Solution : Given: $\angle BAC=90^{\circ}$ and D is mid-point of BC BC would have been the diameter of a circle passing through A, B, and C. So, AD, BD, and CD will be equal to the radius of the circle ⇒ $AD = CD = BD$ Hence, the correct answer is $AD = BD = CD$.
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Question : In $\triangle $ABC, AD$\perp$ BC and AD2 = BD × DC. The measure of $\angle$ BAC is:
Option 1: 60°
Option 2: 75°
Option 3: 90°
Option 4: 45°
Question : The side $BC$ of a triangle $ABC$ is extended to $D$. If $\angle ACD = 120^{\circ}$ and $\angle ABC = \frac{1}{2} \angle CAB$, then the value of $\angle ABC$ is:
Option 1: $80^{\circ}$
Option 2: $40^{\circ}$
Option 3: $60^{\circ}$
Option 4: $20^{\circ}$
Question : $O$ is the orthocentre of triangle $ABC$, and if $\angle BOC = 110^\circ$, then $\angle BAC$ will be:
Option 1: $110^\circ$
Option 2: $70^\circ$
Option 3: $100^\circ$
Option 4: $90^\circ$
Question : $ABC$ is an isosceles triangle with $AB = AC$, The side $BA$ is produced to $D$ such that $AB = AD$. If $\angle ABC = 30^{\circ}$, then $\angle BCD$ is equal to:
Option 1: $45^{\circ}$
Option 2: $90^{\circ}$
Option 3: $30^{\circ}$
Option 4: $60^{\circ}$
Question : In triangle ABC, $\angle$ABC = 15°. D is a point on BC such that AD = BD. What is the measure of $\angle$ADC (in degrees)?
Option 1: 15
Option 2: 30
Option 3: 45
Option 4: 60
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