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Tutorialspoint
52 Views
Given: Sides $AB$ and $AC$ of $\triangle ABC$ are extended to points $P$ and $Q$ respectively. Also, $\angle PBCAB$.Solution:We know that, The sum of the measures of the angles in linear pairs is always $180^o$This implies, $\angle ABC+\angle PBC=180^o$This implies, $\angle ABC=180^o-\angle PBC$In a similar way, we get, $\angle ACB+\angle QCB=180^o$This ... Read More
Tutorialspoint
86 Views
To do:We have to show that in a right-angled triangle, the hypotenuse is the longest side.Solution:Let us consider $ABC$ a right-angled triangleWe know that the sum of the interior angles of the triangle is always $180^o$.This implies, $\angle A+\angle B+\angle C=180^o$$90^o+\angle B+\angle C=180^o$$\angle B+\angle C=180^o-90^o$$\angle B+\angle C=90^o$Now, we have$\angle B+\angle ... Read More
Tutorialspoint
76 Views
Given:$BE$ and $CF$ are two equal altitudes of a triangle $ABC$.To do:By using the RHS congruence rule we have to prove that the triangle $ABC$ is isosceles.Solution:Let us consider $\triangle BEC$ and $\triangle CFB$, We have, $BE$ and $CF$ are two equal altitudes of a triangle $ABC$.This implies, $\angle BEC=\angle ... Read More
Tutorialspoint
53 Views
Given:$ABC$ is an isosceles triangle with $AB=AC$.To do:We have to draw $AP \perp BC$ to show that $\angle B=\angle C$.Solution:Let us consider $\triangle ABP$ and $\triangle ACP$We know that according to the RHS rule if the hypotenuse and one side of a right-angled triangle are equal to the corresponding hypotenuse ... Read More
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Tutorialspoint
23 Views
Given:Two sides $AB$ and $BC$ and median $AM$ of one triangle $ABC$ are respectively equal to sides $PQ$ and $QR$ and median $PN$ of $\triangle PQR$.To do: We have to show that:(i) $\triangle ABM \cong \triangle PQN$(ii) $\triangle ABC \cong \triangle PQR$.Solution:(i) Given, $AM$ is the median of $\triangle ABC$ and $PN$ is ... Read More
Tutorialspoint
48 Views
Given:$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB=AC$.To do:We have to show that:(i) $AD$ bisects $BC$(ii) $AD$ bisects $\angle A$.Solution:Let us consider $\triangle ABD$ and $ACD$Given, that $AD$ is the altitude of $\triangle ABD$ and $ACD$We get, $\angle ADB=\angle ADC=90^o$We have, $AB=AC$Since $AD$ is the common ... Read More
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Tutorialspoint
39 Views
Given:$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$. If $AD$ is extended to intersect $BC$ at $P$.To do:We have to show:(i) $\triangle ABD \cong \triangle ACD$(ii) $\triangle ABP \cong \triangle ACP$(iii) $AP$ ... Read More
Tutorialspoint
50 Views
To do:We have to show that the angles of an equilateral triangle are $60^o$ each.Solution:Let us consider an equilateral triangle $ABC$We have, $AB=BC=AC$ (from fig)We know that The sides opposite to the equal angles are equal.Therefore, $\angle A=\angle B=\angle C$We also know that, The sum of the interior angles of a ... Read More
Tutorialspoint
48 Views
Given:$ABC$ is a right-angled triangle in which $\angle A=90^o$ and $AB=AC$.To do:We have to find $\angle B$ and $\angle C$.Solution:Given, that $AB=AC$We know that, The angles opposite to the equal sides are also equal.This implies, $\angle B = \angle C$ We know that, The sum of the interior angles of a ... Read More
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Tutorialspoint
26 Views
Given:$\triangle ABC$ is an isosceles triangle in which $AB=AC$. Side $BA$ is produced to $D$ such that $AD=AB$. To do:We have to show that $\angle BCD$ is a right angle.Solution:Let us consider $\triangle ABC$, Given, AB = AC We know that, The angles opposite to the equal sides are also equal.This implies, ... Read More