Properties Related to an Isosceles Triangle

The circumcentre of triangle $ABC$is $O$. If angle $OBA=35°$ and $\angle OCA=25°$, then the value of angle $OBC$ is

If we produce the base of an isosceles triangle on both sides then two exterior angles are formed are

In an isosceles triangle $\mathrm{A}\mathrm{B}\mathrm{C},$ with $\mathrm{A}\mathrm{B}=\mathrm{A}\mathrm{C},$ the bisectors of $\mathrm{\angle }\mathrm{B}$ and $\mathrm{\angle }\mathrm{C}$ intersect each other at $\mathrm{O}.$ Join $\mathrm{A}$ to $\mathrm{O}.$ Is $\mathrm{O}\mathrm{B}=\mathrm{O}\mathrm{C}$ (Yes/No).

In the given figure, find $\mathrm{PT}$ given that $l_1\parallel l_2$

If $\Delta ABC$ is right angled at $B$, where $A(5,6,4), B(4,4,1)$ and $C(8,2,x)$ then find the value of $x$ is

In the figure, if $△POQ~△SOR$ and $PQ : RS=$ $1:2,$ then $OP:OS$ is 

Construct an isosceles triangle PQR in which side PQ = PR $= 5$ cm and height PS $= 2.5$cm. 

The triangle is called isosceles when its _____ sides are equal. 

In the figure, $\mathrm{PA}$ is a tangent to the circle, $\mathrm{PBC}$ is secant and $\mathrm{AD}$ bisects angle $\mathrm{BAC}.$ Show that triangle $\mathrm{PAD}$ is an isosceles triangle. Also, show that $\angle \mathrm{CAD}=\frac{1}{2} [\angle \mathrm{PBA}-\angle \mathrm{PAB}].$

Tangent at $\mathrm{P}$ to the circumcircle of triangle $\mathrm{PQR}$ is drawn. If this tangent is parallel to side $\mathrm{QR}$, show that $△\mathrm{PQR}$ is isosceles.
 

$\mathrm{AB}$ is the diameter and $\mathrm{AC}$ is a chord of a circle with centre $\mathrm{O}$ such that angle $\mathrm{BAC}=30°.$ The tangent to the circle at $\mathrm{C}$ intersects $\mathrm{AB}$ produced in $\mathrm{D}.$ Show that $\mathrm{BC}=\mathrm{BD}.$
 

Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.

All isosceles triangles are similar.

If a perpendicular is drawn from the vertex $A$ to the opposite side$BC$, we get $A{D}^2=BD\times DC$, then prove that $∆ABC$ is a right-angled triangle.

In $△ABC$ and $△FDE$, state whether $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$ is true or not? Justify your answer.

In $\Delta ABC$ if $AB=BC$, then

In $△\mathrm{P}\mathrm{Q}\mathrm{R}, \mathrm{\angle }\mathrm{R}=\mathrm{\angle }\mathrm{P}$ and $\mathrm{Q}\mathrm{R}=4 \mathrm{c}\mathrm{m}$ and $\mathrm{P}\mathrm{R}=5 \mathrm{c}\mathrm{m}$. Then the length of $PQ$ is

The perimeters of two similar triangles, $\Delta ABC$ and $\Delta PQR$ are respectively $18 \mathrm{cm}$ and $12 \mathrm{cm}.$ If $PQ=5 \mathrm{cm}$ and $AB=k \mathrm{cm}$, then find $k$ in decimal form.

A vertical stick $1 \mathrm{m}$ long casts a shadow $80 \mathrm{cm}$ long. At the same time a tower casts a shadow $30 \mathrm{m}$ long. If the height of the tower is $k \mathrm{m}$, then find the value of $k$ in decimal form.

In the given figure, $\angle CAB=90°$, $AD\perp BC$ and $△BDA~△BAC$, $AC=75 \mathrm{cm}, AB=1 \mathrm{m}$, $BC=1.25 \mathrm{m}$ and $AD=k \mathrm{cm}$, then the value of $k$ is