Given: ∆ABC, AB = CB BD − altitude to AC P ∈ BD, PB = PC m∠ABC = 48° Find: m∠PCD

Answer: The measure of angle PCD is 42 degree.Explanation:It is given that in ∆ABC,AB = CB  BD − altitude to ACP ∈ BD, PB = PC∠ABC = 48°Since the AB and CB are equal therefore by property of isosceles triangle the angle BAC and angel BCA are equal.According to the angle sum property the the sum of interior angles of a triangle is 180.[tex]\angle ABC+\angle BAC+\angle BCA=180^{\circ}[/tex][tex]48^{\circ}+2\angle BAC=180^{\circ}[/tex][tex]\angle BAC=66^{\circ}[/tex]The altitude BD on AC divides the angle b in two equal parts because triangle ABC is an isosceles triangle.[tex]\angle PBC=\frac{48}{2}=24^{\circ}[/tex]Since PB=PC , so triangle BPC is an isosceles triangle.[tex]\angle PBC=\angle PCB=24^{\circ}[/tex]From the figure it is noticed that,[tex]\angle PCD+\angle PCB=\angle BCA[/tex][tex]\angle PCD+24^{\circ}=66^{\circ}[/tex][tex]\angle PCD=66^{\circ}-24^{\circ}[/tex][tex]\angle PCD=42^{\circ}[/tex]Therefore, the measure of angle PCD is 42 degree.