Goa Board Solutions for Chapter: Triangles, Exercise 1: EXERCISE

$ABCD$ is a quadrilateral in which $AD=BC$ and $\mathrm{\angle }DAB=\mathrm{\angle }CBA$ (see given figure). Prove that $BD=AC$.

$ABCD$ is a quadrilateral in which $AD=BC$ and $\mathrm{\angle }DAB=\mathrm{\angle }CBA$ (see given figure). Prove that $\mathrm{\angle }ABD=\mathrm{\angle }BAC$.

Line $l$ is the bisector of an angle $\mathrm{\angle }\mathrm{A}\mathrm{}$ and $\mathrm{B}$ is any point on $l.$ $\mathrm{B}\mathrm{P}$ and $\mathrm{B}\mathrm{Q}$ are perpendiculars from $\mathrm{B}$ to the arms of $\mathrm{\angle }\mathrm{A}$ (see given figure). Show that
$\mathrm{B}\mathrm{P}=\mathrm{B}\mathrm{Q}$ or $\mathrm{B}$ is equidistant from the arms of $\mathrm{\angle }\mathrm{A}.$

$\mathrm{A}\mathrm{B}$ is a line segment and $\mathrm{P}$ is its mid-point. $\mathrm{D}$ and $\mathrm{E}$ are points on the same side of $\mathrm{A}\mathrm{B}$ such that $\mathrm{\angle }\mathrm{B}\mathrm{A}\mathrm{D}=\mathrm{\angle }\mathrm{A}\mathrm{B}\mathrm{E}$ and $\mathrm{\angle }\mathrm{E}\mathrm{P}\mathrm{A}=\mathrm{\angle }\mathrm{D}\mathrm{P}\mathrm{B}$ as shown in figure. Show that $\mathrm{AD}=\mathrm{BE}$.

In right triangle $\mathrm{A}\mathrm{B}\mathrm{C},$ right-angled at $\mathrm{C}, \mathrm{M}$ is the midpoint of hypotenuse $\mathrm{A}\mathrm{B}. \mathrm{C}$ is joined to $\mathrm{M}$ and produced to a point $\mathrm{D}$ such that $\mathrm{D}\mathrm{M}=\mathrm{C}\mathrm{M}.$ Point $\mathrm{D}$ is joined to point $\mathrm{B}$ as shown in figure. Show that $∆\mathrm{AMC}\cong ∆\mathrm{BMD}$.
 
 

In right triangle $\mathrm{A}\mathrm{B}\mathrm{C},$ right angled at $\mathrm{C}, \mathrm{M}$ is the midpoint of hypotenuse $\mathrm{A}\mathrm{B}. \mathrm{C}$ is joined to $\mathrm{M}$ and produced to a point $\mathrm{D}$ such that $\mathrm{D}\mathrm{M}=\mathrm{C}\mathrm{M}.$ Point $\mathrm{D}$ is joined to point $\mathrm{B}$ as shown in figure. Show that $\mathrm{\angle }\mathrm{D}\mathrm{B}\mathrm{C}$ is a right angle.

 

In right triangle $\mathrm{A}\mathrm{B}\mathrm{C},$ right-angled at $\mathrm{C}, \mathrm{M}$ is the midpoint of hypotenuse $\mathrm{A}\mathrm{B}. \mathrm{C}$ is joined to $\mathrm{M}$ and produced to a point $\mathrm{D}$ such that $\mathrm{D}\mathrm{M}=\mathrm{C}\mathrm{M}.$ Point $\mathrm{D}$ is joined to point $\mathrm{B}$ as shown in figure. Show that $∆\mathrm{D}\mathrm{B}\mathrm{C}\cong ∆\mathrm{A}\mathrm{C}\mathrm{B}$.

 

In right triangle $\mathrm{A}\mathrm{B}\mathrm{C},$ right-angled at $\mathrm{C}, \mathrm{M}$ is the midpoint of hypotenuse $\mathrm{A}\mathrm{B}. \mathrm{C}$ is joined to $\mathrm{M}$ and produced to a point $\mathrm{D}$ such that $\mathrm{D}\mathrm{M}=\mathrm{C}\mathrm{M}.$ Point $\mathrm{D}$ is joined to point $\mathrm{B}$ as shown in figure. Show that $\mathrm{C}\mathrm{M}=\frac{1}{2}\mathrm{A}\mathrm{B}$.