NCERT Class 10
D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC Show that CA square = CB.CD.
05
Oct
D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC Show that CA square = CB.CD. D is a point on the side BC of a triangle ABC such that angle ADC = angle BAC Show that CA square = CB.CD. October 5, 2020 Category: Chapter [...]
Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ and PR and median PM of triangle PQR. Show that triangle ABC ~ triangle PQR .
05
Oct
Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ and PR and median PM of triangle PQR. Show that triangle ABC ~ triangle PQR . Sides AB and BC and Median AD of a triangle ABC are proportional to sides PQ and PR and median PM of triangle [...]
In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF.
05
Oct
In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that Δ ABD ~ Δ ECF. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ [...]
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, Show that: (i) CD/GH = AC/FG (ii) ΔDCB ∼ ΔHGE (iii) ΔDCA ∼ ΔHGF
05
Oct
CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ∼ ΔFEG, Show that: (i) CD/GH = AC/FG (ii) ΔDCB ∼ ΔHGE (iii) ΔDCA ∼ ΔHGF CD and GH are respectively the bisectors of ∠ACB and [...]
In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) △ABC∼△AMP (ii) PA/CA = MP/BC.
04
Oct
In Fig., ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) △ABC∼△AMP (ii) PA/CA = MP/BC. ABC and AMP are two right triangles In Fig. right angled at B and M respectively. Prove that: (i) △ABC∼△AMP (ii) PA/CA = MP/BC. October 4, 2020 Category: Chapter 6 - [...]
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB.
04
Oct
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ∼ ΔCFB. October 4, 2020 Category: Chapter 6 - [...]
In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ∼ ΔCDP (ii) ΔABD ∼ ΔCBE (iii) ΔAEP ∼ ΔADB (v) ΔPDC ∼ ΔBEC
04
Oct
In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ∼ ΔCDP (ii) ΔABD ∼ ΔCBE (iii) ΔAEP ∼ ΔADB (v) ΔPDC ∼ ΔBEC altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ∼ ΔCDP (ii) ΔABD [...]
In figure, if ΔABE ≅ ΔACD, show that ΔADE ∼ ΔABC.
04
Oct
In figure, if ΔABE ≅ ΔACD, show that ΔADE ∼ ΔABC. if ΔABE ≅ ΔACD In figure show that ΔADE ∼ ΔABC. October 4, 2020 Category: Chapter 6 - Triangles , Maths , NCERT Class 10 ,
S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS Show that triangle RPQ ~ triangle RTS.
04
Oct
S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS Show that triangle RPQ ~ triangle RTS. S and T are points on sides PR and QR of triangle PQR such that angle P = angle RTS Show that triangle RPQ ~ triangle RTS. October [...]
In the figure, = QR /QS = QT/PR and ∠1 = ∠ 2 then prove that ∆PQS ~ ∆TQR .
04
Oct
In the figure, = QR /QS = QT/PR and ∠1 = ∠ 2 then prove that ∆PQS ~ ∆TQR . = QR /QS = QT/PR and ∠1 = ∠ 2 then prove that ∆PQS ~ ∆TQR . In the figure October 4, 2020 Category: Chapter 6 - Triangles , Maths , NCERT Class 10 ,