Given: ∆ABC, AB = CB BD − median to AC E∈ AB ,F∈ BC AE = CF Prove: △ADE ≅ △CDF ΔBDE ≅ ΔBDF
1 answer:
Answer:
1) By SAS theorem, ΔADE≅ΔCDF
2) By SSS theorem, ΔBDE≅ΔBDF
Step-by-step explanation:
Consider isosceles triangle ABC (see diagram).
1. In triangles ADE and CDF:
- AD≅DC (since BD is median, then it divides side AC in two congruent parts);
- AE≅CF (given);
- ∠A≅∠C (triangle ABC is isosceles, then angles adjacent to the base are congruent).
By SAS theorem, ΔADE≅ΔCDF.
2. In triangles BDE and BDF:
- side BD is common;
- DE≅DF (ΔADE≅ΔCDF, then congruent triangles have congruent corresponding sides);
- BE≅FB (triangle ABC is isosceles, AB≅BC, AE≅CF, then BE=AB-AE, FB=BC-CF).
Be SSS theorem, ΔBDE≅ΔBDF.
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Let:
Vj = Speed of jane
Vm = Speed of mike
d = distance
t = time
Jane travels 7 mph less than 2 times as fast as Mike, therefore:
Vj = 2Vm - 7
Remeber:
distance = speed*time
Distance traveled by mike:
d=Vm*t = Vm*6
Distance traveled by jane:
d + 198 = Vj*6
where:
Vj = 2Vm - 7
d + 198 = (2Vm-7)*6
Now, let:
d=Vm*6 (1)
d + 198 = (2Vm-7)*6 (2)
Replace (1) into (2)
6Vm + 198 = 12Vm - 42
Subtract 6Vm from both sides:
6Vm + 198 - 6Vm = 12Vm - 42 - 6Vm
198 = 6Vm - 42
Add 42 to both sides:
198 + 42 = 6Vm - 42 + 42
240 = 6Vm
Divide both sides by 6:
240/6 = 6Vm/6
40 = Vm
Vm = 40mph
Replace Vm into this equation Vj = 2Vm - 7 :
Vj = 2(40) - 7 = 80 - 7 = 73mph
